Question: Solve for $y$, $ -\dfrac{8}{3y^3} = \dfrac{5y - 9}{3y^3} - \dfrac{10}{3y^3} $
If we multiply both sides of the equation by $3y^3$ , we get: $ -8 = 5y - 9 - 10$ $ -8 = 5y - 19$ $ 11 = 5y $ $ y = \dfrac{11}{5}$